Problem: Example

Input

8 5
1 2
6 5
6 4
1 3
4 7


Output

11

Solution (in CPP):

Solution:
#include<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<list>
#include<bitset>
#include<stack>
#include<memory>
#include<numeric>
using namespace std;
using namespace std::chrono;
typedef long long int llint;
typedef long double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase
/*cout<<fixed<<setprecision(20);cin.tie(0);ios::sync_with_stdio(false);*/
const llint mod=1000000007 ;
const llint big=2.19e15+1;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-10;
const long double epsA=1e-4;
template <class T,class U>bool mineq(T& a,U b){if(a>b){a=b;return true;}return false;}
template <class T,class U>bool maxeq(T& a,U b){if(a<b){a=b;return true;}return false;}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){if(a==0){return b;}return a/gcd(a,b)*b;}
template<class T> void SO(T& ve){sort(ve.begin(),ve.end());}
template<class T> void REV(T& ve){reverse(ve.begin(),ve.end());}
template<class T>llint LBI(const vector<T>&ar,T in){return lower_bound(ar.begin(),ar.end(),in)-ar.begin();}
template<class T>llint UBI(const vector<T>&ar,T in){return upper_bound(ar.begin(),ar.end(),in)-ar.begin();}

int main(void){
	cout<<fixed<<setprecision(20);cin.tie(0);ios::sync_with_stdio(false);
	llint n,e,i;cin>>n>>e;
	vector<vector<int>>go(n);
	for(i=0;i<e;i++){
		
		int a,b;cin>>a>>b;a--;b--;
		go[a].pub(b);go[b].pub(a);
		
	}
	
	vector<int>pa(n,-1);
	queue<int>que;
	int base=0;
	static int kaz[100001]={0};
	for(i=0;i<n;i++){
		
		if(pa[i]!=-1){continue;}
		
		pa[i]=0;que.push(i);
		int g=0,k=0;
		while(que.size()){
			int t=que.front();que.pop();
			int z=pa[t];
			if(z){g++;}
			else{k++;}
			for(auto it:go[t]){
				//cerr<<it<<endl;
				if(pa[it]==-1){que.push(it);pa[it]=1-z;}
				if(pa[it]==z){cout<<-1<<endl;return 0;}
			}
		}
		//cerr<<"g="<<g<<"k="<<k<<endl;
		base+=min(g,k);
		kaz[abs(g-k)]++;//偏り
	}
	bitset<100001>bits;bits[base]=1;
	for(i=1;i<=n;i++){
		if(kaz[i]==0){continue;}
		if(kaz[i]>2){
			int t=(kaz[i]-2)/2;
			kaz[i]-=t*2;kaz[i+i]+=t;
		}
		if(kaz[i]>0){bits|=(bits<<i);}
		if(kaz[i]>1){bits|=(bits<<i);}
	}
	llint ans=0;
	for(i=0;i<=n;i++){
		if(bits[i]){maxeq(ans,i*(n-i));}
	}
	cout<<ans-e<<endl;
	//gの合計をできるだけn/2-baseに近づける
	
	return 0;
}